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O lol. I never knew that. Well then, I do acknowledge his proficiency with the subject then.

@genius2014 Why are you automatically assuming that you and Drongoski are the only competent mathematicians on this site? People such as ohexploitable and Bored of Fail 1,2,3,...,n, and many others are as capable as you guys are. Each year members who have been helped by this site return to help...

Thanks Drongoski. I understand where I went wrong. And for what its worth I do appreciate people like Drongoski taking time out and helping other people. Money shouldn't be an incentive to give others a hand. So thank you Drongoski.

a) A flag pole 9m high stands on top of a support. The top if the support is 16m above the eye-level of the observer. If the flagpole subtends an angle, theta, at the eye of the observer who is standing x metres away from the support: It looks like this: http://i51.tinypic.com/28lb7ly.png...

Listen through my computer with headphones, as computer is laptop --> crap speakers. Headphones are Sennheiser CX400II, quite nice with a great bass response. Anywhere else I use my iPod.

+1. I have a Vista. Sexy it is.

You mean Fairy Tail? The one where all the females have big boobs and the character is a dragon slayer? If so then yes.

Hot damn guys!!! Appreciate the effort! That's some nice music there. I especially like 'Thank You' by Home Made Kazoku, 'Feel Like Dance', 'Tokyo Jihen' and 'Sukima Switch'. Unfortunately I'm not a big fan of visual kei, like 'The Gazette' cause it just sounds a bit weird. But the last 2...

My fav is 5318008. That's because when you put it in a calculator, upside-down it spells boobies. Raj from Big Bang ftw.

Trying to pick-up japanese as a casual pursuit. Already watching tons of anime, and thought that songs would also assist me in my pursuit. So any suggestions of good japanese songs? Any genre except for heavy rock/death metal is fine. After a certain point, screaming in Japanese to me is...

Normanhurst Boys

Binders + loose leaf is so much more convenient. At the end of a unit, you just file away the loose leafs and start afresh. The sheets don't end up clogging up exercise books, and your bag is a bit lighter. BUT, it also kinda depends on what type of binder as well. 2 hole binders ruin the...

Those days were great... I started off with Time Spiral in year 9, had a white weenie deck consisting of knights. Then I switched to a Kithkin deck in Lorwyn. Then Red/Green agro/burn/rush in Shadowmoor. Stopped playing after that. Sorry for all the jargon but i just had to let it rip :) It...

The solution is in the complex field, which is 4U. I thought this was 2U revision?

Ooooohhhhhh... I see... Thanks guys for the quick response! So basically every time you sub in a value of n another value of 'I' pops up, and so on until there is only a singe value left. And reduction formulae are just formulae that symbolise the general solution of a sine function raised to...

Somewhere in the middle-end of Motors, where me and my friends are trying to get everyone to accept the theory that eddy currents explains everything in the physics, as 'eddy currents' was the only response a substitute gave us to all of our questions.

cos(2A) = 1-2sin^2(A) sin^2(A) = (1-cos(2A))/2 If A = x - (pi/4), then 2A = 2x - (pi/2) Then you substitute the integrand with 1-cos(2x-(pi/2)) . 0.5 And then you integrate it.

What's the purpose of reduction formulae in integration? I tried doing it and failed miserably. I was thinking that maybe I would be more proficient at it if I interpreted the concept in a different way.

Why would you even need to achieve an exceptional result in the SC? 0-0.5 hours for that. I spent 30 mins after school and got all 6's, granted they were low 90s. For HSC I would say 2 hours after school, cause I'm doing that right now and it feels better than doing nothing and it's just right...

∠PRM = ∠MKQ (int.∠ equal to opp. ext. angle of cyclic quads.) PR = RM (tangents drawn to common ext. point (R) are equal in length) therefore ΔPRM is isos. MK = QK (radii egual in length) therefore ΔMKQ is isos. Hence, base angles of both triangles are equal as ∠PRM = ∠MKQ. Hence, ΔMKQ///ΔPRM...